Question

what is the \axe\ description of the tetrachloroiodide (ici₄⁻) anion?

Explanation:

Step1: Determine the central atom and bonding/non - bonding regions

The central atom in \( \text{ICl}_4^- \) is Iodine (I). First, we calculate the number of valence electrons around the central atom. Iodine is in group 17, so it has 7 valence electrons. Each Cl contributes 7 valence electrons, and there are 4 Cl atoms, so \( 4\times7 = 28 \) valence electrons from Cl. The anion has a - 1 charge, so we add 1 more electron. The total number of valence electrons is \( 7+28 + 1=36 \).

The formula for the number of electron groups (bonding + non - bonding) around the central atom: For a molecule/ion with formula \( AX_nE_m \), where \( A \) is the central atom, \( X \) is the number of bonded atoms, and \( E \) is the number of lone pairs on the central atom.

The number of bonding pairs (X): I is bonded to 4 Cl atoms, so \( n = 4 \).

To find the number of lone pairs (m) on the central atom: The number of electron groups around I is calculated as follows. The number of electron groups is \( \frac{36}{8}=4.5 \)? No, we use the formula for the number of electron groups: \( \text{Number of electron groups}=\frac{\text{Valence electrons of central atom}+\text{number of bonded atoms}\times1+\text{charge (for anion +, for cation -)}}{2} \)

Wait, a better way: The central atom I has 7 valence electrons, it forms 4 single bonds with Cl (each bond uses 1 electron from I and 1 from Cl), so the number of electrons used in bonding is \( 4\times2 = 8 \) (but I contributes 4 electrons for bonding). The remaining electrons on I: \( 7 - 4+1 \) (the +1 from the charge) \( = 4 \), which is 2 lone pairs? Wait, no. Let's use the VSEPR formula. The formula for the steric number (number of electron domains) is \( \text{Steric number}=\text{number of bonded atoms}+\text{number of lone pairs on central atom} \)

For \( \text{ICl}_4^- \), the central atom I:

The number of valence electrons for I: 7

Number of electrons from Cl: 4 (each Cl donates 1 electron for bonding)

Charge: +1 (so we add 1 electron)

Total electrons around I: \( 7+4 + 1=12 \)

Number of electron pairs: \( \frac{12}{2}=6 \)

Number of bonding pairs (X): 4 (since it's bonded to 4 Cl atoms)

Number of lone pairs (E): \( 6 - 4=2 \)? Wait, no, that's wrong. Wait, the correct way:

The Lewis structure of \( \text{ICl}_4^- \):

Iodine is the central atom. It forms 4 single bonds with Cl atoms. Each single bond has 2 electrons. So 4 bonds account for \( 4\times2 = 8 \) electrons. The remaining electrons: total valence electrons \( = 7 (I)+4\times7(Cl)+1(charge)=7 + 28+1 = 36 \). The electrons used in bonding: \( 4\times2 = 8 \), so the remaining electrons are \( 36-8 = 28 \). These remaining electrons are distributed as lone pairs on Cl and I. Each Cl needs 6 more electrons (since it has 1 bond, 2 electrons, and needs 8), so 4 Cl atoms need \( 4\times6 = 24 \) electrons. So the remaining electrons on I: \( 28 - 24=4 \), which is 2 lone pairs. Wait, no, the steric number (number of electron domains) is the number of bonding domains plus the number of lone pair domains.

The steric number for I in \( \text{ICl}_4^- \): The number of bonding domains (X) is 4 (bonded to 4 Cl), and the number of lone pair domains (E) is 2? Wait, no, let's check the VSEPR theory. The formula for \( \text{ICl}_4^- \):

The central atom I has 7 valence electrons, 4 Cl atoms (each with 7 valence electrons), and a - 1 charge (so +1 electron). The total number of valence electrons is \( 7+4\times7 + 1=36 \).

The number of electron groups (steric number) around I: We arrange the electron groups. The number of bonding pairs (X) is 4 (I - Cl bonds), and the number of lone pairs (E) is 2? Wait, no, actually, the correct steric number calculation: For a central atom, the number of electron domains is \( \frac{\text{valence electrons of central atom}+\text{number of atoms bonded to it}-\text{charge (if cation) + charge (if anion)}}{2} \)

So for I: \( \frac{7 + 4-(- 1)}{2}=\frac{7 + 4+1}{2}=\frac{12}{2}=6 \). So there are 6 electron domains. Out of these, 4 are bonding domains (X = 4) and 2 are lone pair domains (E = 2)? Wait, no, that can't be. Wait, no, the formula for \( \text{ICl}_4^- \) is actually having a square planar geometry. In square planar geometry, the steric number is 6 (octahedral electron geometry), with 4 bonding pairs and 2 lone pairs? Wait, no, square planar has 4 bonding pairs and 2 lone pairs? No, square planar has 4 bonding pairs and 2 lone pairs in the octahedral electron geometry (the two lone pairs are opposite each other). Wait, but let's check the AXE notation. The AXE notation is \( AX_nE_m \), where \( n \) is the number of bonded atoms, and \( m \) is the number of lone pairs on the central atom.

Wait, maybe I made a mistake. Let's look up the structure of \( \text{ICl}_4^- \). The central I atom has 4 bonded Cl atoms and 2 lone pairs? No, wait, the correct structure: Iodine in \( \text{ICl}_4^- \) has 7 valence electrons, plus 1 from the charge, total 8? No, no. Wait, the correct calculation:

The number of valence electrons for I: 7

Number of Cl atoms: 4, each contributes 1 electron for bonding, so 4 electrons.

Charge: - 1, so +1 electron.

Total electrons around I: \( 7+4 + 1=12 \), which is 6 electron pairs.

Number of bonding pairs: 4 (since it's bonded to 4 Cl), so number of lone pairs: \( 6 - 4=2 \). So the AXE notation is \( AX_4E_2 \)? Wait, no, that's for an octahedral electron geometry with square planar molecular geometry. But wait, maybe I was wrong. Wait, let's check the actual AXE notation for \( \text{ICl}_4^- \). The central atom I has 4 bonded atoms (X = 4) and 2 lone pairs (E = 2)? No, wait, no. Wait, the correct steric number: The formula for the number of electron domains (steric number) is \( \text{number of bonded atoms}+\text{number of lone pairs} \). For \( \text{ICl}_4^- \), the central I atom:

The Lewis structure: I is bonded to 4 Cl atoms. Each Cl has 3 lone pairs. The I atom: let's count the electrons. Each I - Cl bond is a single bond (2 electrons). So 4 bonds: 8 electrons. The I atom has 7 valence electrons, plus 1 from the charge, so 8 electrons. Wait, 8 electrons around I: 4 bonding pairs (8 electrons) and 0 lone pairs? No, that can't be. Wait, I think I messed up the valence electron calculation.

Wait, the correct way: The total number of valence electrons in \( \text{ICl}_4^- \) is \( \text{Valence electrons of I}+\text{Valence electrons of 4 Cl}+\text{charge} \)

Valence electrons of I: 7

Valence electrons of Cl: 7, 4 Cl: \( 4\times7 = 28 \)

Charge: - 1, so we add 1 electron.

Total: \( 7+28 + 1=36 \)

Now, we distribute these electrons. The central I atom is bonded to 4 Cl atoms. Each I - Cl bond is a single bond (2 electrons), so 4 bonds use \( 4\times2 = 8 \) electrons.

Remaining electrons: \( 36-8 = 28 \)

These remaining electrons are distributed as lone pairs on the Cl atoms and the I atom. Each Cl atom needs 6 more electrons (since it has 1 bond, 2 electrons, and needs 8 to satisfy the octet), so 4 Cl atoms need \( 4\times6 = 24 \) electrons.

Remaining electrons after filling Cl atoms: \( 28 - 24 = 4 \), which are 2 lone pairs on the I atom.

So the central I atom has 4 bonding pairs (X = 4) and 2 lone pairs (E = 2). Wait, but 4 bonding pairs (8 electrons) and 2 lone pairs (4 electrons) make 12 electrons around I, which is more than 8. But iodine is in the fifth period, so it can have an expanded octet. So that's okay.

So the AXE notation is \( AX_4E_2 \)? Wait, no, that's not right. Wait, maybe I made a mistake. Let's check the VSEPR theory. The formula \( AX_nE_m \), where \( n \) is the number of bonded atoms, and \( m \) is the number of lone pairs on the central atom.

For \( \text{ICl}_4^- \), the central atom is I. The number of bonded atoms (X) is 4, and the number of lone pairs (E) is 2? No, wait, no. Wait, the correct structure: Iodine in \( \text{ICl}_4^- \) has a steric number of 6 (4 bonding, 2 non - bonding), so the AXE notation is \( AX_4E_2 \)? But that seems to be for a square planar molecular geometry (with octahedral electron geometry). But maybe I was wrong. Wait, let's check an example. The \( \text{ICl}_4^- \) ion: the central I atom has 4 Cl atoms bonded to it and 2 lone pairs. So the AXE notation is \( AX_4E_2 \)? No, wait, no. Wait, maybe the correct AXE notation is \( AX_4E_2 \)? Wait, no, I think I made a mistake. Let's think again.

Wait, the number of valence electrons on the central atom: I has 7, plus 1 from the charge, total 8? No, the charge is - 1, so we add 1 electron to the whole ion, not just the central atom. So the total valence electrons are 7 (I) + 4*7 (Cl) + 1 (charge) = 36, as before.

The central I atom: we use the formula for the number of electron domains (steric number) = (number of valence electrons of central atom + number of atoms bonded to it - charge (if cation) + charge (if anion))/2

So (7 + 4 - (-1))/2=(7 + 4+1)/2 = 12/2 = 6. So steric number is 6. Number of bonded atoms (X) is 4, so number of lone pairs (E) is 6 - 4 = 2. So the AXE notation is \( AX_4E_2 \).

Wait, but maybe I was wrong. Let's check a reference. The \( \text{ICl}_4^- \) ion has a square planar geometry, with the central I atom having 4 bonding pairs and 2 lone pairs. So the AXE notation is \( AX_4E_2 \).

Step2: Confirm the AXE notation

So in the AXE notation \( AX_nE_m \), \( A \) is the central atom, \( X \) is the number of bonded atoms, and \( E \) is the number of lone pairs. For \( \text{ICl}_4^- \), \( n = 4 \) (4 Cl atoms bonded to I) and \( m = 2 \) (2 lone pairs on I). So the AXE notation is \( AX_4E_2 \).

Answer:

\( AX_4E_2 \)